Question

Вычислить определенный интеграл:

Answer 1

$1)quad intlimits _{frac{1}{sqrt{e}}}^{e}frac{dx}{x}=ln|x|limits |^{e}_{frac{1}{sqrt{e}}}=lne-lnfrac{1}{sqrt{e}}=1-(ln1-lnsqrt{e})=\\=1-(0-frac{1}{2})=frac{1}{2}\\2)quad intlimits^0_{-frac{pi}{2}} {sin(frac{pi}{4}-3x)} , dx =-frac{1}{3}cdot (-cos(frac{pi}{4}-3x))|_{-frac{pi}{2}}^0=\\=frac{1}{3}left (cosfrac{pi}{4}-cos(frac{pi}{4}+frac{3pi}{2})right )=frac{1}{3}left (frac{sqrt2}{2}-cosfrac{7pi}{4}right )=$

$=frac{1}{3}left (frac{sqrt2}{2}-frac{sqrt2}{2}right )=0$

$3)quad int limits _0^1, arctgx, dx=[, u=arctgx,; du=frac{dx}{1+x^2}, ,, dv=dx; ,; v=x]=\\=xcdot arctgx|_0^{frac{pi}{4}}-int limits _{0}^{frac{pi}{4}}, frac{x, dx}{1+x^2} =frac{pi}{4}cdot arctgfrac{pi}{4}-frac{1}{2}cdot int limits _0^{frac{pi}{4}}frac{2x, dx}{1+x^2}=\\=frac{pi}{4}cdot arctgfrac{pi}{4}-frac{1}{2}cdot ln|1+x^2||_0^{frac{pi}{4}}=\\=frac{pi}{4}cdot arctgfrac{pi}{4}-frac{1}{2}(ln(1+frac{pi ^2}{16})-ln1)=$

$=frac{pi}{4}cdot arctgfrac{pi}{4}-frac{1}{2}ln(1+frac{pi ^2}{16})$